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Given this system: enter image description here

I need to show the $\mathcal Z$-transform of $y[n]$ as a function of the $\mathcal Z$-transform of $x[n]$.

Now I know that for downsampling alone:

$$Y(z) = \frac1M\sum_{m=0}^{M-1} X\left(e^\frac{-j2\pi{m}}M\cdot z^\frac 1M\right)$$

And that for upsampling alone:

$$Y(z) = X\left(z^M\right)$$

My problem here is that I don't quite understand how upsampling would affect the downsampling equation. Say we call the point between downsampling and upsampling $w[n]$, we get:

\begin{align}Y(z) &= W\left(z^M\right)\\ W(z) &= \frac1M\sum_{m=0}^{M-1} X\left(e^\frac{-j2\pi{m}}M\cdot z^\frac 1M\right)\end{align}

I'm not sure if I will be getting:

$$Y(z) = \frac1M\sum_{m=0}^{M-1} X\left(e^\frac{-j2\pi{m}}M\cdot z\right)$$

Or:

$$Y(z) = \frac1M\sum_{m=0}^{M-1} X\left(e^{-j2\pi{m}}\cdot z\right)$$

Or even something else.

I'll be very happy for some clarification on this subject because I'm very baffled by it and I cannot find a solution anywhere, thank you.

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You will get

$Y(z) = \frac1M\sum_{m=0}^{M-1} X(e^\frac{-j2\pi{m}}M*z)$


If you were to upsample first and then downsample you would get

$Y(z) = \frac1M\sum_{m=0}^{M-1} X(e^{-j2\pi{m}}*z)$

which just simplifies to (since $e^{-j2\pi{m}}=1$ for $m\in\mathbb{Z}$)

$\frac1M\sum_{m=0}^{M-1} X(e^{-j2\pi{m}}*z) = \frac1M\sum_{m=0}^{M-1} X(z) = X(z)$

Basically, if you upsample and downsample you can perfectly reconstruct the signal but if you downsample and then upsample, you are aliasing in the summation over rotations in the z-domain of the input.

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  • $\begingroup$ What is your set $\mathbb I$ ? $\endgroup$ – Gilles Feb 3 '16 at 11:03
  • $\begingroup$ It is the set of all integers $\endgroup$ – cimarron Feb 3 '16 at 16:19
  • $\begingroup$ Ah, maybe you meant $\mathbb Z$ ? It's the first time I see $\mathbb I$ for integers. $\endgroup$ – Gilles Feb 3 '16 at 16:57
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    $\begingroup$ yes, indeed, updated the post $\endgroup$ – cimarron Feb 3 '16 at 17:02

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